I have not tried all the way though, but here is a way to start. Let for any . Note that . Now, . It is easy to see that does not depend on itself, but only on , so the last sum is equal to . Both of these factors are easy to compute.

first, here is a different problem:
how many permutations in are there which has fix points?
second, to compute , we can consider the derivative of . expand the bracket, find the derivative and then let x=1, we get.